Question

If the solubility of a gas is 8.4 g/L at 404 kPa pressure, what is the solubility of the gas when the pressure is 101 kPa? Please show work.

Answer 1

at P = 101 KPa ⇒ S = 2.1 g/L

Step-by-step explanation:

Henry's law:

• C = K*P

∴ K: Henry's constant

∴ C is the concentration ≡ Solubiluty (S)

∴ S = (8.4 g/L)*(Kg/1000 g)*(1000 L/m³) = 8.4 Kg/m³

∴ P = (404 KPa)*(1000 Pa/KPa) = 404000 Pa

∴ Pa = Kg/ms²

⇒ S = K(404000 Pa) = 8.4 Kg/m³

⇒ K = 2.079 E-5 s²/m².......the henry's constant for this gas

at P = 101 KPa = 101000 Pa

⇒ S = (2.079 E-5 s²/m²)*(101000 Kg/ms²)

⇒ S = 2.1 Kg/m³ = 2.1 g/L